Talk:Boolean prime ideal theorem
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"Apparently, the ultrafilter lemma also implies BPI, such that both statements are equivalent -- please confirm if this is known to you." I managed to work out a rather convoluted proof of this, showing that ultrafilter lemma-->compactness theorem-->BPI for free Boolean algebras-->BPI. But I get the feeling there ought to be a more direct proof, and although I was very careful, I might have tacitly used some aspect of the axiom of choice at some point in my proof. --—Preceding unsigned comment added by 188.8.131.52 (talk • contribs)
- The compactness theorem (even for sentential logic) directly implies the BPI:
- Fix a Boolean algebra B. For every b in B let pb be a propositional variable. Let T be the set of all formulas of the following form:
- not pb iff pnot b
- (pb and pc) iff pb and c
- (pb or pc) iff pb or c
- where "not", "and", "or" on the left side is a logical symbol, and "not", "and", "or" on the right side are the boolean operations. T is easily seen to be finitely consistent, and every truth assignment to the propositional variables that satisfies T will induce an ultrafilter on B.
- --Aleph4 22:20, 2 May 2006 (UTC)
"the Boolean prime ideal theorem is often taken as an axiom of set theory". Is it really? Ever? That sounds like there are (many) people who actually work in the set theory "ZF + BPI". I doubt that. Of course there are few people who investigate the strength of BPI, but I doubt that they consider BPI an "axiom of set theory".
--Aleph4 18:21, 24 September 2005 (UTC)
merged ultrafilter lemma
Weak BPI for Boolean algebras
Weak and strong PIT are equivalent for Boolean algebras via quotient algebras. The same should hold for distributive lattices, but I am not absolutely certain. TexD 19:12, 23 October 2006 (UTC)